Two chords AB and CD of circle whose centre is `O`, meet at the point `P and /_AOC = 50^@, /_BOD = 40^@`. Then the value of `/_BPD `is

Options

A. `60^@`
B. `40^@`
C. `45^@`
D. `75^@`

Answer

To solve the problem, we need to find the value of angle \( \angle BPD \) given that \( \angle AOC = 50^\circ \) and \( \angle BOD = 40^\circ \).

### Step-by-step Solution:

1. **Identify the Angles at the Center:**
- We know that \( \angle AOC = 50^\circ \) and \( \angle BOD = 40^\circ \).

2. **Calculate the Angles at the Circumference:**
- The angle at the circumference is half the angle at the center for the same arc.
- Therefore, for chord \( AC \):
\[ \angle APC = \frac{1}{2} \times \angle AOC = \frac{1}{2} \times 50^\circ = 25^\circ \]
- For chord \( BD \):
\[ \angle BPD = \frac{1}{2} \times \angle BOD = \frac{1}{2} \times 40^\circ = 20^\circ \]

3. **Use the Exterior Angle Theorem:**
- In triangle \( BPC \), \( \angle BPD \) is the exterior angle.
- According to the exterior angle theorem:
\[ \angle BPD = \angle APC + \angle BPC \]
- Therefore, we can express \( \angle BPD \) as:
\[ \angle BPD = \angle APC + \angle BPC = 25^\circ + 20^\circ = 45^\circ \]

4. **Conclusion:**
- The value of \( \angle BPD \) is \( 45^\circ \).

### Final Answer:
\[ \angle BPD = 45^\circ \]